\(\int \frac {1}{(a+b x^2)^{5/2}} \, dx\) [93]

Optimal result

Integrand size = 11, antiderivative size = 39 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {x}{3 a \left (a+b x^2\right )^{3/2}}+\frac {2 x}{3 a^2 \sqrt {a+b x^2}} \]

[Out]

1/3*x/a/(b*x^2+a)^(3/2)+2/3*x/a^2/(b*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {198, 197} \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}} \]

[In]

Int[(a + b*x^2)^(-5/2),x]

[Out]

x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x}{3 a \left (a+b x^2\right )^{3/2}}+\frac {2 \int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a} \\ & = \frac {x}{3 a \left (a+b x^2\right )^{3/2}}+\frac {2 x}{3 a^2 \sqrt {a+b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {3 a x+2 b x^3}{3 a^2 \left (a+b x^2\right )^{3/2}} \]

[In]

Integrate[(a + b*x^2)^(-5/2),x]

[Out]

(3*a*x + 2*b*x^3)/(3*a^2*(a + b*x^2)^(3/2))

Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {x \left (2 b \,x^{2}+3 a \right )}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}\)	\(26\)
trager	\(\frac {x \left (2 b \,x^{2}+3 a \right )}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}\)	\(26\)
pseudoelliptic	\(\frac {x \left (2 b \,x^{2}+3 a \right )}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}\)	\(26\)
default	\(\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\)	\(32\)

[In]

int(1/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*x*(2*b*x^2+3*a)/(b*x^2+a)^(3/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, b x^{3} + 3 \, a x\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} \]

[In]

integrate(1/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*b*x^3 + 3*a*x)*sqrt(b*x^2 + a)/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (32) = 64\).

Time = 0.50 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.44 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {3 a x}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{3}}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} \]

[In]

integrate(1/(b*x**2+a)**(5/2),x)

[Out]

3*a*x/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a)) + 2*b*x**3/(3*a**(7/2)*sqrt(1 + b
*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {2 \, x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} \]

[In]

integrate(1/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

2/3*x/(sqrt(b*x^2 + a)*a^2) + 1/3*x/((b*x^2 + a)^(3/2)*a)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {x {\left (\frac {2 \, b x^{2}}{a^{2}} + \frac {3}{a}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/3*x*(2*b*x^2/a^2 + 3/a)/(b*x^2 + a)^(3/2)

Mupad [B] (verification not implemented)

Time = 4.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {2\,x\,\left (b\,x^2+a\right )+a\,x}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}} \]

[In]

int(1/(a + b*x^2)^(5/2),x)

[Out]

(2*x*(a + b*x^2) + a*x)/(3*a^2*(a + b*x^2)^(3/2))